Is g Continuous if g^-1(O) is Open for All Open Sets O?

[kathrynag]

Homework Statement

Let g be defined on all of R. If A is a subset of R, define the set g^-1(A) by
g^-1(A)={x in R : g(x) in A}.
Show that g is continuous iff g^-1(O) is open whenever O contained in R is an open set.

Homework Equations

The Attempt at a Solution

well g^-1(O) means g(O) is in A.
Let g be continuous and O be an open subset of R.
Then |x-c|<delta and |g(x)-g(c)|<epsilon

Now I get stuck

 

[jbunniii]

A good start would be to state the definition of continuity you are using.

 

[kathrynag]

A function f:A--->R is continuous at a point c in A if, for all epsilon epsilon>0, there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.

 

[ns2675]

Just appeal to the definitions.

First suppose g^-1(O) is open for each open subset O of R. Fix a real number x, and let epsilon be given. The neighborhood N centered at g(x) with radius epsilon is an open set, so the inverse image of N under g is an open set containing x. Thus there is a neighborhood V of x with radius delta such that V is a subset of g^-1(N); you should be able to finish up the forward direction from here.

The reverse implication plays out similarly. If g is continuous at x, then there exists a delta such that whenever y is less than delta apart from x, g(y) is in a neighborhood of g(x) with radius epsilon. You can fill in the details, but this pretty much shows that the inverse image of this epsilon-neighborhood is open. Since the inverse image of any open set is the union of the inverse images of the neighborhoods whose union is that open set, it follows that the inverse image under g of any open set is itself open.

 

[kathrynag]

I don't follow this part:
You can fill in the details, but this pretty much shows that the inverse image of this epsilon-neighborhood is open. Since the inverse image of any open set is the union of the inverse images of the neighborhoods whose union is that open set, it follows that the inverse image under g of any open set is itself open.

 

[ns2675]

Suppose g is continuous, and O is an open set in R. Suppose x is in the pre-image of O under g; that is, g(x) is in O. Then there is a neighborhood N centered at g(x) with radius epsilon such that N is a subset of O. Since g is continuous at x, there is a positive number delta such that g(y) is in N (hence in O) just as soon as the distance between x and y is less than delta. In other words, there is a neighborhood V centered at x with radius delta which is a subset of the pre-image of O under g.